Let $x \in [0,1[$ and $\theta$ a real. Show that $$\frac{1}{1-x} - \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right) \geq \frac{x(1-\cos\theta)}{(1-x)\left((1-x)^2 + 2x(1-\cos\theta)\right)}.$$ Deduce that if $x \geq \frac{1}{2}$ then $$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1-\cos\theta}{6(1-x)^3}\right) \quad \text{or} \quad \left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{3(1-x)}\right).$$ For this last result, distinguish two cases according to the relative values of $x(1-\cos\theta)$ and $(1-x)^2$.
Let $x \in [0,1[$ and $\theta$ a real. Show that
$$\frac{1}{1-x} - \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right) \geq \frac{x(1-\cos\theta)}{(1-x)\left((1-x)^2 + 2x(1-\cos\theta)\right)}.$$
Deduce that if $x \geq \frac{1}{2}$ then
$$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1-\cos\theta}{6(1-x)^3}\right) \quad \text{or} \quad \left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{3(1-x)}\right).$$
For this last result, distinguish two cases according to the relative values of $x(1-\cos\theta)$ and $(1-x)^2$.